Question

Two boys having masses $m_1=4\text{kg}$ and $m_2=2\text{kg}$ are seating on a seesaw of length $r=1\text{m}$ as shown in figure. Find the net torque due to weight of boys, acting about the midpoint $O$.
(Take $g=10{\text{m/s}}^2$ and anti-clock wise rotation as $+ve$)

• A. $-10\text{N-m}$
• B. $20\text{N-m}$
• C. $10\text{N-m}$
• D. $-20\text{N-m}$

Answer 1

The correct option is C $10\text{N-m}$



Let torque due to $m_{1}g$ and $m_{2}g$ are ${\tau }_1$ and ${\tau }_2$ respectively.

We know;
$\tau =R\times F$
Taking anti-clock wise rotation $(+ve)$
$\therefore {\tau }_1=\frac{r}{2}\times \left(m_{1}g\right)=\frac{1}{2}\times (4\times 10)=20\text{N-m}$
and
$\therefore {\tau }_2=-\frac{r}{2}\times \left(m_{2}g\right)=-\frac{1}{2}\times (2\times 10)=-10\text{N-m}$

We know,
${\tau }_{net}={\tau }_1+{\tau }_2=(20-10)\text{N-m}=10\text{N-m}$